Uno to Mega and where is the error for the interrupt that isn't delared?

Question

I want to read a Omron E6B2-CWZ6C Incremental Rotary Encoder w/Index (360P/R). A couple questions, I found out that an Uno can only handle 2 interrupt pins and I think I need 3 for the Index. What do I need to do to the following code to get it to run on a Mega which I believe can handle 3 interrupt pins?

Also I am getting a 'outAChange' was not declared in this scope. What am I missing here.

I will be using the data from the encoder value to run a NeoPixel 60 for a directional mast/antenna.

Here is the code I am trying to use:

// Wiring connections for my encoder:
// Brown : VCC = 5 to 24VDC
// Blue: 0V(Common) = GND
// Shield: GND
// Black: outA = Digital pin 2
// White: outB = Digital pin 3
// Orange: outZ = Digital pin 4 (Index)
// With these outA/outB/outZ connections and the interrupt code below
// clockwise rotation gives positive encoder counts
const int outA = 2;
const int outB = 3;
const int outZ = 4;
volatile long encoder = 0; // declare volatile since modified by interrupt routines
float encoder_save = 0;
void setup(){ // set encoder pins
pinMode(outA, INPUT_PULLUP);
pinMode(outB, INPUT_PULLUP);
pinMode(outZ, INPUT_PULLUP); //Index
// attach interrupts to pins
attachInterrupt(digitalPinToInterrupt(outA), outAChange, CHANGE);
attachInterrupt(digitalPinToInterrupt(outB), outBChange, CHANGE);
attachInterrupt(digitalPinToInterrupt(outZ), outZChange, CHANGE);
// initialize serial
Serial.begin(57600);
}
void loop(){ 
if (encoder_save != encoder) // only print if new value
encoder_save = encoder;
// If Index(Z) reads low reset encoder to 0
if (digitalRead(outZ) == LOW)
{
 encoder = 0;
}
}
{
Serial.println((encoder * 360)/360);
}
}
// interrupt routines
void outAChange(){
// when outA changes, outA==outB means negative direction
encoder += digitalRead(outA) == digitalRead(outB) ? -1 : 1;
if (encoder < 0)
{
 encoder = (encoder + 360); 
}
}
void outBChange(){
// when outB changes, outA==outB means positive direction
encoder += digitalRead(outA) == digitalRead(outB) ? 1 : -1;
if (encoder >= 360)
{
 encoder = (encoder - 360);
}
}

Answer 1

Your brackets are all messed up. If you pass your code through an autoformatter (such as the one included in the IDE) you end up with this:

// Wiring connections for my encoder:
// Brown : VCC = 5 to 24VDC
// Blue: 0V(Common) = GND
// Shield: GND
// Black: outA = Digital pin 2
// White: outB = Digital pin 3
// Orange: outZ = Digital pin 4 (Index)
// With these outA/outB/outZ connections and the interrupt code below
// clockwise rotation gives positive encoder counts
const int outA = 2;
const int outB = 3;
const int outZ = 4;
volatile long encoder = 0; // declare volatile since modified by interrupt routines
float encoder_save = 0;
void setup() { // set encoder pins
 pinMode(outA, INPUT_PULLUP);
 pinMode(outB, INPUT_PULLUP);
 pinMode(outZ, INPUT_PULLUP); //Index
// attach interrupts to pins
 attachInterrupt(digitalPinToInterrupt(outA), outAChange, CHANGE);
 attachInterrupt(digitalPinToInterrupt(outB), outBChange, CHANGE);
 attachInterrupt(digitalPinToInterrupt(outZ), outZChange, CHANGE);
// initialize serial
 Serial.begin(57600);
}
void loop() {
 if (encoder_save != encoder) { // only print if new value
 encoder_save = encoder;
 }
// If Index(Z) reads low reset encoder to 0
 if (digitalRead(outZ) == LOW) {
 encoder = 0;
 }
}
{ << WHAT
 Serial.println((encoder * 360) / 360); << IS
} << THIS
} << CODE?
// interrupt routines
void outAChange() {
// when outA changes, outA==outB means negative direction
 encoder += digitalRead(outA) == digitalRead(outB) ? -1 : 1;
 if (encoder < 0) {
 encoder = (encoder + 360);
 }
}
void outBChange() {
// when outB changes, outA==outB means positive direction
 encoder += digitalRead(outA) == digitalRead(outB) ? 1 : -1;
 if (encoder >= 360) {
 encoder = (encoder - 360);
 }
}

As you can see, after the end of your loop() function, you have an extraneous block of code with a bunch of random brackets.

You need to decide where in your program that should actually be and move it there, deleting those extra brackets.

• Bkukuk62: I could use a little more help than that. I am really new at coding. thanks.
• Majenko: I cannot help you. Only you know where you want that bit of code.